Question

Find three rational numbers between \[1.0100100010000....\] and \[1.012303003000.....\]

Answer 1

Hint: Any number which has a string of digits which repeats itself after the non-repeating part, after a decimal point is a rational number. Now, find any three numbers for which the digits start repeating itself after a non-repeating part after the decimal point, keeping in mind two conditions i) the numbers are greater than \[1.0100100010000....\]as well as ii) they are lesser than \[1.012303003000.....\]

Complete step-by-step answer:
A rational number is any real number which can be expressed in the form $\dfrac{\text{p}}{\text{q}}$, where p and q are two integers which are co-prime to each other. Now, any periodic decimal number is a rational number.
Suppose, a decimal number is of the form $\text{x}\text{.}{{\text{p}}_{1}}{{\text{p}}_{2}}......{{\text{p}}_{\text{k}}}{{\text{q}}_{1}}{{\text{q}}_{2}}......{{\text{q}}_{\text{l}}}{{\text{q}}_{1}}{{\text{q}}_{2}}......{{\text{q}}_{\text{l}}}......$ where, ${{\text{p}}_{1}}{{\text{p}}_{2}}......{{\text{p}}_{\text{k}}}$ is the non-repetitive part where ${{\text{q}}_{1}}{{\text{q}}_{2}}......{{\text{q}}_{\text{l}}}$is the repetitive periodic part after the decimal point. It can be expressed in the form (x + r) where, r = ${{\text{p}}_{1}}{{\text{p}}_{2}}......{{\text{p}}_{\text{k}}}{{\text{q}}_{1}}{{\text{q}}_{2}}......{{\text{q}}_{\text{l}}}{{\text{q}}_{1}}{{\text{q}}_{2}}......{{\text{q}}_{\text{l}}}......$
Let, m = ${{\text{p}}_{1}}{{\text{p}}_{2}}......{{\text{p}}_{\text{k}}}$ and n = ${{\text{p}}_{1}}{{\text{p}}_{2}}......{{\text{p}}_{\text{k}}}{{\text{q}}_{1}}{{\text{q}}_{2}}......{{\text{q}}_{\text{l}}}$
Now, we can say that,
\[\begin{align}
 & {{10}^{\text{k}}}\text{ x r = (}{{\text{p}}_{1}}{{\text{p}}_{2}}......{{\text{p}}_{\text{k}}}).{{\text{q}}_{1}}{{\text{q}}_{2}}......{{\text{q}}_{\text{l}}}{{\text{q}}_{1}}{{\text{q}}_{2}}......{{\text{q}}_{\text{l}}}....\text{ (A)} \\
 & \text{1}{{\text{0}}^{\text{k + l}}}\text{ x r = (}{{\text{p}}_{1}}{{\text{p}}_{2}}......{{\text{p}}_{\text{k}}}{{\text{q}}_{1}}{{\text{q}}_{2}}......{{\text{q}}_{\text{l}}}).{{\text{q}}_{1}}{{\text{q}}_{2}}......{{\text{q}}_{\text{l}}}.....\text{ (B)} \\
\end{align}\]
Subtracting equation (A) from equation (B), we get,
$\begin{align}
 & \left( {{10}^{\text{k+l}}}\text{ }-\text{ 1}{{\text{0}}^{\text{k}}} \right)\text{ x r = }{{\text{p}}_{1}}{{\text{p}}_{2}}......{{\text{p}}_{\text{k}}}{{\text{q}}_{1}}{{\text{q}}_{2}}......{{\text{q}}_{\text{l}}}\text{ }-\text{ }{{\text{p}}_{1}}{{\text{p}}_{2}}......{{\text{p}}_{\text{k}}} \\
 & \text{ = n - m} \\
 & \therefore \text{ r = }\dfrac{\text{n }-\text{ m}}{{{10}^{\text{k+l}}}\text{ }-\text{ 1}{{\text{0}}^{\text{k}}}} \\
\end{align}$

Hence, by the definition of a rational number, r is a rational number.
Thus, the original number (x + r) is also a rational number.
Hence, any number which has a string of digits which repeats itself after the non-repeating part, after a decimal point is a rational number.

Thus, three rational numbers between \[1.0100100010000....\] and \[1.012303003000.....\]are:
i) 1.01111111.......
ii) 1.011222222.....
iii) 1.012222222......

Note: The three rational numbers in the answer are randomly chosen. Any three rational numbers other than these three are also valid answers, with the condition keepin in mind that those three numbers should be in between \[1.0100100010000....\] and \[1.012303003000.....\]